This amazing website
hopes to provide you with everything you need to
know for your
maths GCSE, nothing more and nothing less! These clear, comprehensive and fun
notes have already helped people just like you to achieve their true potential
at GCSE.

At mathsmad.com we
know that the best way to learn maths is to practice. Please take advantage
of our completely free 'quick
questions' section after learning from our extensive
notes
pages.

These notes were written by James Wawrzynski, a medical
student at the University of Cambridge. He achieved an A* at GCSE maths and an A at A-level.

If you find our website
useful, please recommend it to your friends at school or tell us what you think
by writing to info@mathsmad.com
Below are some quick links to the many GCSE topics explained on this site...

Factorisation (4 types) -There are four
different ways of factorising expressions

1- Take out a common factor

x2 + 3x + x

=x (x + 3 + 1)

=x (x + 4)

In the example above, each of the three parts that are added
together are divisible by "x", hence we divide each by "x", put the whole thing
in brackets and put "x" in front of the bracket.

2- Quadratic factorisation

x2 + 4x - 5

In this case, x2
and 4x are both divisible by x but 5 is not hence we can't use method 1. We have
to factorise in the form (x + p)(x + q). 'p' and 'q' are two numbers that
together add to 4 and multiply to -5.

Quadratics that you have to factorise will
always be in the form ax2 + bx
+c. 'p' and 'q' must add to b and multiply to c.

3- Harder quadratic
factorisation

20x2 - 7x - 6

You will notice that in method 2, 'a' in ax2
+ bx +cwas equal to 1. Method three is
used when the number in front of x2
is greater than 1. The problem with this is that we don't know whether to write
(20x + p)(x + q) or (10x +p)(10x + q) etc...

1- Always write (20x + p)(20x + q) / 20

2- p and q must add to b and multiply to c*a in ax2
+ bx +c.

Qu) But in method 2 they had to multiply to c, not c*a (???)

Ans) Actually, they have to add to b and multiply to c*a in both, except in
method 2, 'a' was 1 so 'c*a' was the same as 'c'

4- Difference of two squares (expressions in the form
ax2 - c where a and c are
both square numbers)

4x2 - 9

The square root both parts of it to get 2x
and 3

write (2x + 3)(2x - 3)

This means that when you multiply it out
using FOIL (first, outer, inner, last) the first and last terms give us 4x2 - 9 and the outer and inner are
-6x + 6x and therefore cancel.

Completing the square

This is a slightly different form of factorisation

x2 + 10x + 25

25 is "the square"

10 is "twice the square root of the
square"

1- always write (x+p)(x+p) where p is half of
"twice the square root of the square" , so in this case we write
(x+5)(x+5)

2- multiply it out and see what you get. In
this case we get x2 + 10x + 25 so
(x+5)(x+5) is correct. The reason why it works out is because c in ax2 + bx +
c was exactly the square of half of b

What if c is not exactly the
square of half of b?

x2 + 10x + 22

1- do exactly the same... Write
(x+p)(x+p) where p is half of b, so in this case we write (x+5)(x+5)

2- multiply it out and see what you get. In
this case we get x2 + 10x + 25. But the
original question said x2 + 10x + 22 so our answer is +3 too big
therefore we have to change it to (x+5)(x+5) - 3.

NB) The usual way of writing these answers is
(x+5)2 - 3
instead of (x+5)(x+5) - 3.

Completing the square can be used to solve quadratic equations.

x2 + 10x + 22 = 0

(x+5)2
- 3 = 0

(x+5)2
= 3

x+5 =
+-√3

x = -5 +-√3

x = - 3.27 or x = -6.73
(all answers must be to 3 s.f. always!)

The Quadratic formula- alternative
to completing the square

1- write down the equation you have to solve
e.g. 3x2 + 6x = 7

2-re write in the form ax2
+ bx + c = 0 3x2 + 6x - 7 = 0

3- a= 3, b=
6, c= -7

4- Plug these into the
quadratic formula x=[-b +-√(b2
- 4ac)] / 2a

5- The answer for this example turns out to
be x = 0.826 or x = -2.83

Exam tip) Any time the question says "give
your answer to a suitable degree of accuracy" you know that the answer will be a
very long decimal and so the original expression does not factorise so you have
to use the formula!

Fractions in equations

E.g. 1

[x+2]/6 +3 = [4x]/5

Multiply both sides by 6 and then by 5 to
get rid of the denominators

5x + 10 + 90 = 24x

E.g. 2

2/[n+2] +3/[n-1] = 5

Multiply both sides by (n+2) and then by
(n-1) to get rid of the denominators

If you feel you have any quality content that you would
like to share, email the owner of this site directly on
info@mathsmad.com

If you have any comments or suggestions, email
info@mathsmad.com.
Your opinion is highly valued and some comments may make their way onto these
pages!

Topics on
this site include: Algebra, Geometry, Indices, Irrational numbers, Circle
theorem, Simultaneous equations, Equations of Circles, Proportionality,
Rearranging Equations, Graphical solution of equations (using graphs to solve
equations), Histograms, Sampling, Probability, Similar Solids, Proof, Accuracy
and Error, Trigonometric functions, Sin (Sine) and cosine rule, Transformations
and Vectors. Although this website is based on the
AQA syllabus spec. A, many topics for other boards such as EdExcell and the IGCSE
overlap with topics here. We hope this site is one of the best on the web,
however mathsmad cannot guarantee the accuracy or completeness of the
information on this website. All pages of this site are the copyright of the owner and may not be reproduced without written permission from the owner. This statment supercedes any other information that may be provided on this website in the case of contradiction.